...By quantizing we the get the following Hamiltonian operator

$$\hat{H}=\sum_{\mathbf{k}}\hbar \omega(\mathbf{k})\left(\hat{n}(\mathbf{k})+\frac{1}{2} \right)$$ where $\hat{n}(\mathbf{k})=\hat{a}^{\dagger}(\mathbf{k})\hat{a}(\mathbf{k})$ is the number operator of oscillator mode $\mathbf{k}$ with eigenvalues $n_{\mathbf{k}}=0,1,2,\dots$.

Using the quantum canonical ensemble show that the internal energy $E(T)$ is given by>

$$E(T)=\langle H \rangle = E_0 + \sum_{\mathbf{k}}\frac{\hbar \omega(\mathbf{k})}{e^{\beta\hbar \omega(\mathbf{k})}-1}$$

where $E_0$ is the sum of ground state energies of all the oscillators.

I started this by calculating the partition function

$$\begin{align} Z &= \sum_{\Gamma}e^{-\beta \mathcal{H}(\Gamma)} \\ &= \sum_{\Gamma}e^{-\beta (\sum_{\mathbf{k}}\hbar \omega(\hat{n}(\mathbf{k})+\frac{1}{2}))} \end{align}$$ ($\Gamma$ is a microstate of the system)

but I cannot see the thought process behind evaluating these, particularly with respect to the summations. This is a common problem I have found.

I would then go on to use $E=-\frac{\partial \ln Z}{\partial \beta}$

  • $\begingroup$ Note all the oscillators are independent $\endgroup$
    – Trajan
    Apr 16, 2015 at 20:17

2 Answers 2


Quantum mechanically the general expression you want for the partition function is $$ Z = \mathrm{Tr} \left( \mathrm{e}^{-\beta H} \right),$$ where $\mathrm{Tr}$ means the trace (i.e. sum over micro-states). Now you can use the fact that the modes are independent, so that quantum Boltzmann operator $\mathrm{e}^{-\beta H}$ factorises into a product. This means that you can evaluate the trace over each oscillator mode separately: $$ Z = \mathrm{Tr} \left( \mathrm{e}^{-\beta H} \right) = \mathrm{Tr} \left( \prod_\mathbf{k}\mathrm{e}^{-\beta H_\mathbf{k}}\right) = \prod_\mathbf{k} \mathrm{Tr}_\mathbf{k} \left( \mathrm{e}^{-\beta H_\mathbf{k}}\right) = \prod_\mathbf{k} Z_\mathbf{k} $$ where $\mathrm{Tr}_\mathbf{k}$ means the trace over only the Hilbert space of mode $\mathbf{k}$, and $$H_\mathbf{k} = \hbar\omega(\mathbf{k})\left(\hat{n}(\mathbf{k}) + \frac{1}{2}\right).$$ Now $\mathrm{Tr}_\mathbf{k}$ means simply averaging over all the possible states in the Hilbert space, which you might as well choose to be the eigenstates of the number operator $\hat{n}(\mathbf{k})\lvert m_\mathbf{k}\rangle= m_\mathbf{k} \lvert m_\mathbf{k}\rangle$, with $m_\mathbf{k} = 0,1,2,\ldots$. So you have to evaluate $$ Z_\mathbf{k} = \mathrm{Tr}_\mathbf{k} \left( \mathrm{e}^{-\beta H_\mathbf{k}}\right) = \sum_{m_\mathbf{k}=0}^\infty \langle m_\mathbf{k} \rvert \mathrm{e}^{-\beta H_\mathbf{k}} \lvert m_\mathbf{k} \rangle. $$

  • 1
    $\begingroup$ Following on from your answer I get $Z_\mathbf{k} = \frac{e^{-\hbar\omega(\mathbf{k})/2}}{1-e^{-\hbar\omega(\mathbf{k})m_{\mathbf{k}}}}$. So $Z=\prod_{\mathbf{k}} \frac{e^{-\hbar\omega(\mathbf{k})/2}}{1-e^{-\hbar\omega(\mathbf{k})m_{\mathbf{k}}}}$. I dont see how the answer will follow $\endgroup$
    – Trajan
    Apr 17, 2015 at 9:13
  • 1
    $\begingroup$ @sandstone You're answer cannot depend on $m_\mathbf{k}$ (i.e. the number of bosons corresponding to a micro-state of mode $\mathbf{k}$) since you have already summed over this variable. You need the result for a geometric series. You are on the right track, keep at it. $\endgroup$ Apr 17, 2015 at 10:55
  • $\begingroup$ I see that it should have been, $Z_\mathbf{k} = \frac{e^{-\beta\hbar\omega(\mathbf{k})/2}}{1-e^{-\beta\hbar\omega(\mathbf{k})}}$, I had misapplied the geometric series . I cannot still see how to evaluate the product $\endgroup$
    – Trajan
    Apr 17, 2015 at 11:50
  • $\begingroup$ @sandstone You do not need to evaluate the product. What you need is to calculate $-\partial \ln Z/\partial\beta = -\frac{1}{Z}\partial Z/\partial \beta$, right? $\endgroup$ Apr 17, 2015 at 12:12
  • $\begingroup$ Oh that makes much more sense $\endgroup$
    – Trajan
    Apr 17, 2015 at 12:15

My original answer was incorrect and Mark gave a good answer above.

Here is a slightly different approach by considering the total energy of the system. Say we have $\alpha$ total harmonic oscillators. We can write the total energy of the whole system as

$$ E = \sum_\alpha \frac{1}{2}\hbar\omega_\alpha + \sum_\alpha n_\alpha \hbar \omega_\alpha .$$

For each oscillator $\alpha$, $n_\alpha = 0,1,2,3...$

The first term in the expression for the total energy is the sum of ground state energies of all the oscillators, $E_0$.

$$ \Rightarrow E_0 = \sum_\alpha \frac{1}{2}\hbar\omega_\alpha $$

The we can write the partition function as the sum of all possible subsytems that have this total energy.

$$ Z = \sum_{\{n_\alpha\}} e^{-\sum_\alpha \frac{1}{2}\hbar\omega_\alpha /kT} e^{-\sum_\alpha n_\alpha \hbar \omega_\alpha /kT}$$

This simplifies to...

$$ Z = e^{-E_0/kT} \prod_\alpha \left( 1+e^{-\hbar\omega_\alpha /kT}+ e^{-2\hbar\omega_\alpha /kT}+e^{-3\hbar\omega_\alpha /kT}+... \right) $$

Rewriting the series of exponential terms gives us...

$$ Z = e^{-E_0/kT} \prod_\alpha \left( \frac{1}{1-e^{-\hbar\omega_\alpha /kT}}\right)$$

From here you can use the formula $U = -\frac{\partial \ln Z}{\partial\beta}$ to find the internal energy.


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